Answer
$\quad f'(2) = 0.5\\\quad f'(16) = 0.0625\\\quad f'(-3) = -0.\bar3$
Work Step by Step
$f(x) = \ln|x|$
$\implies f'(x) = \lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}$
$\implies f'(x) = \lim\limits_{h \to 0} \frac{\ln|x+h|-\ln|x|}{h}$
$\implies f'(2) = 0.5\\\quad f'(16) = 0.0625\\\quad f'(-3) = -0.\bar3$
(These values have been obtained using a graphing calculator, image attached [$h$ has been reduced to a value approximately close to zero to obtain results])
